# Is the application of Le Chatelier's principle wrong in this case?

In the reaction: $$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$ Suppose the system has 3 moles of $\ce{N2}$, 1 mole of $\ce{H2}$ and 1 mole of $\ce{NH3}$ at equilibrium. $Q_\chi=K_\chi=8.33$ and $Q_p=K_p$.

Now I add 5 moles of $\ce{N2}$ in the container by keeping (i) pressure and (ii) volume constant (temperature constant in both).

According to Le Chatelier's principle:

• If pressure is constant: equilibrium shifts to backward direction.
• If volume is constant: equilibrium shifts to forward direction.

But if I look at the equilibrium in terms of mole fraction, $Q_\chi$ will be the same in both the cases and hence equilibrium should shift in the same direction i.e. in backward direction.

Why do they predict different equilibrium states?

[$K_\chi$ is equilibrium constant in terms of mole fraction, $K_p$ is equilibrium constant in terms of partial pressures].

Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature. It depends on total pressure as well as temperature. Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed. In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.

Le Chatelier's principle says that, for a system at equilibrium, a change in temperature, pressure or concentration of ONE of the components will tend to shift to counteract the change.

Consider first case ii, constant volume. If additional nitrogen is introduced into this volume, the concentration (pressure) of nitrogen will increase, while all other concentrations (partial pressures) remain the same. By Le Chatelier's principle, this single change tends to drive the reaction toward a higher concentration of NH3 (forward, not backward).

Keeping total pressure constant (case i) requires expanding the volume by a factor of two (3+1+1 moles increases to 3+1+1+5 moles). This changes ALL of the concentrations (pressures).

The concentration of N2 increases by a factor of (8/2)/3 = 4/3. This is 8 moles in twice the volume vs 3 moles in the original volume, thus favoring the forward direction. The effect of H2 is a factor of 8 in the backward direction (1/2 cubed) or 1/8 forward and the effect of decreased NH3 is a factor of 4 in the forward direction (1/2 squared). The overall effect will be 4/3 x 1/8 x 4 = 16/24 = 2/3 in the forward direction; i.e., the addition of extra N2 and volume expansion now favors the reverse direction, compared to the original situation direction.

Keeping the total pressure constant violates Le Chatelier's principle by changing more than one variable (concentrations of reactants and product). However, it is just a complication that can be handled by looking at the equilibrium constant; it's nice that the temperature remains the same.

• In the constant volume case, the concentration of N2 does not increase by a factor of six. Before, there were 3 moles in a given volume, now there are 8 moles in twice the volume, so a ratio of 3 to 4, not 1 to 6. Also, adding the effects does not make sense, you have to multiply them.
– Karsten
Oct 21, 2021 at 20:24
• @Karsten Theis: You are quite right! Thank you. I think I must have been using wrong numbers. But the multiplication rather than addition of effects is one for me to remember. Oct 22, 2021 at 2:52

For this reaction:

$$\Delta n=-2$$

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(i) Adding an inert gas at constant temperature and pressure implies an increase in total volume, so:

$$V_1

Also at constant temperature and pressure, we have:

$$Q_c=K_c*\left(\frac{V_1}{V_2}\right)^{\Delta n}$$

Let:

$$\Upsilon=\frac{V_1}{V_2}$$

$$Q_c=K_c*\Upsilon^{-2}$$

$$\ce{V1 < V2 -> \Upsilon < 1->Q_c > K_c}$$

So equilibrium shifts to the left.

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(ii) Adding an inert gas at constant temperature and volume implies an increase in total pressure, so:

$$P_1

Also at constant temperature and volume, we have:

$$Q_p=K_p*\left(\frac{P_2}{P_1}\right)^{\Delta n}$$

Let:

$$\rho=\frac{P_2}{P_1}$$

$$Q_p=K_p*\rho^{-2}$$

$$\ce{P1 < P2 ->\rho > 1->Q_c < K_c}$$

So equilibrium shifts to the right.